# KBCC The History of Math and Ancient Greek Mathematics Worksheet

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Anen88

Mathematics

CUNY Kingsborough Community College

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2} to form a triangular number we begin with 1

We say 1 is the first triangular number, to form the next we add gnomon 2:

So 3 is the second consecutive triangular number
The sum of the two consecutive triangular numbers
=1+3=4
To proof 4 as a square number by picture

3} if the nth odd number is 2n-1
So we prove that
1+3+…+ (2n-3) + (2n-1) =𝑛2
It can be proved using arithmetic progression
Sum of an A.P with n terms, difference and last term
𝑛

With the formula 𝑠𝑛 = 2 (𝑎 + 𝑙)
=𝑡1=𝑎 , 𝑡2=𝑎+2𝑑… 𝑡𝑛=𝑎+(𝑛−1)𝑑
𝑛
2

=𝑠𝑛 = (𝑎 + 𝑎 + (𝑛 − 1)𝑑
𝑛

=𝑠𝑛= 2 (𝑎 + 𝑡𝑛 )
Applying this to the series
𝑛

S=2 (1 + 2𝑛 − 1)
𝑛

=2 (2𝑛) = 𝑛2
4}
5} using the first triangular number 1

Multiplied by 8 =1*8=8

=
Nine is a square number: proof

6}Triangular numbers =1,3,6,10,15,21,28,36,45,55,66,78,91,105,120,136,153,171,190…etc.
16=10+6
25=10+15
39=21+15+3
56=55+1
69=45+21+3
150=105+45
185=120+55+10
287=190+91+6
1
𝑎

1

1

7a} − = −

1
𝑏

Finding the LCM at both sides
ℎ−𝑎

= 𝑎ℎ =

𝑏−ℎ
𝑏ℎ

Cross multiply
=𝑏ℎ ∗ (ℎ − 𝑎) = 𝑎ℎ ∗ (𝑏 − ℎ)
ℎ−𝑎

𝑎ℎ

ℎ−𝑎

𝑎

=𝑏−ℎ = 𝑏ℎ
=𝑏−ℎ = 𝑏

Therefore h satisfies the value of harmonic means of a and b in this relation
2𝑎𝑏

7b} ℎ = 𝑎+𝑏

Cross multiply
=h (a+b) =2ab
= ha+hb=2ab
This proves that h does not satisfy its value for harmonic mean of a and b
8} the area of the four bold triangles is equal to the area of the large, oblique square minus the small
square in the center.
9}
1

10} Area of the trapezoid (EBCD) = 𝐴 = 2 (𝑎 + 𝑏)(𝑎 + 𝑏)
=

(𝑎+𝑏)2
2
1

1

1

Area of the the three triangles (ABC+EAD+EBA) = 𝐴 = 2 (𝑎 ∗ 𝑏) + 2 (𝑐 ∗ 𝑐) + 2 (𝑎 ∗ 𝑏)
=

𝑎𝑏
2

+

𝑐2
2

+

𝑎𝑏
2

= 𝑎𝑏 +

𝑐2
2

Since area of trapezoid=area of 3 triangles
=

(𝑎+𝑏)2
2

𝑐2

= 𝑎𝑏+ 2

=(𝑎 + 𝑏)2 = 2𝑎𝑏 + 𝑐 2
=𝑎2 + 𝑏 2 + 2𝑎𝑏 = 2𝑎𝑏 + 𝑐 2
=𝑎2 + 𝑏 2 = 𝑐 2 𝑝𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑒𝑎𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
12a} (3, 4, 5)
=any other Pythagorean triple are common multiples of the above Pythagorean theorem
To get another valid Pythagorean triple (a,b,c)
(k*a ,k*b ,k*c) where k is a positive integer
E.g. Let k =2
Then (2*3,2*4,2*5)
=(6,8,10) which is a Pythagorean triple
:. The triples cannot be consecutive positive integers
Hence (3,4,5) is the only Pythagorean triple

12b} 𝑥 = 2𝑛 + 1, 𝑦 = 2𝑛2 + 2𝑛, 𝑧 = 2𝑛2 + 2𝑛 + 1 𝑓𝑜𝑟 𝑛 ≥ 1 …equation 1
Pythagoras arrived at the above solution from the relation:
(2𝑘 − 1) + (𝑘 − 1)2 = 𝑘 2 …equation 2
And then searching for those k for which 2k-1 is a perfect square, I.e., 2k1=𝑚2 (𝑠𝑖𝑛𝑐𝑒 𝑚2 𝑖𝑠 𝑜𝑑𝑑, 𝑚 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑜𝑑𝑑). 𝑡ℎ𝑖𝑠 𝑔𝑖𝑣𝑒𝑠:
=𝑘 =

𝑚2 +1
2

𝑎𝑛𝑑 𝑘 − 1 =

𝑚2 −1
2

Thus the relation (2) can be written as:
= 𝑚2 +

(𝑚2 −1)2
2

=

(𝑚2 +1)

2

2

From which it is clear that (1) is satisfied with:
𝑚2 �...

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